# Another spanning set question.

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• Apr 3rd 2009, 06:13 PM
scorpion007
Another spanning set question.
I keep getting inconsistent results with this one. I can't find a single vector that belongs to both sets. Is the question correctly formed?

Question:
Two subspaces S and T of R^3 are spanned by {(1,1,-2), (-1,1,0) } and { (1,1,0), (0,-11,1)} respectively. Find a non-zero vector X that belongs to $\displaystyle S\cap T$.

This is what I did:

We need $\displaystyle X=a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1)$.

So,
$\displaystyle a-b=c$
$\displaystyle a+b=c-11d$
$\displaystyle -2a=d$

Right so far?

From here I can't get any consistent solution for a, b such that $\displaystyle a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1)$.
• Apr 3rd 2009, 08:48 PM
scorpion007
I don't know if I'm doing this right, but if I solve for a and b in terms of c and d, I get:

$\displaystyle -2a=d \Longrightarrow \boxed{a = -\frac{1}{2}d}$

$\displaystyle a-b=c \Rightarrow \left(-\frac{1}{2}d\right)-b=c \Longrightarrow \boxed{b=-c-\frac{1}{2}d}$

But then if I solve the middle equation for b (to check), I get:

$\displaystyle a+b=c-11d \Rightarrow \left(-\frac{1}{2}d\right)+b=c-11d \Longrightarrow \boxed{b=c-\frac{21}{2}d}$

The last two b's are inconsistent! Have I done something wrong?
• Apr 3rd 2009, 09:29 PM
mr fantastic
Quote:

Originally Posted by scorpion007
I keep getting inconsistent results with this one. I can't find a single vector that belongs to both sets. Is the question correctly formed?

Question:
Two subspaces S and T of R^3 are spanned by {(1,1,-2), (-1,1,0) } and { (1,1,0), (0,-11,1)} respectively. Find a non-zero vector X that belongs to $\displaystyle S\cap T$.

This is what I did:

We need $\displaystyle X=a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1)$.

So,
$\displaystyle a-b=c$
$\displaystyle a+b=c-11d$
$\displaystyle -2a=d$

Right so far?

From here I can't get any consistent solution for a, b such that $\displaystyle a(1,1,-2)+b(-1,1,0) = c(1,1,0) + d(0,-11,1)$.

Let $\displaystyle a = \lambda$.

Then $\displaystyle d = -2 \lambda$.

So $\displaystyle c = -10 \lambda$ and $\displaystyle b = 11 \lambda$.

So one particular solution is $\displaystyle a = 1, ~ b = 11, ~ c = -10, ~ d = -2$.

So one possible vector is $\displaystyle (1, 1, -2) + 11 (-1, 1, 0) = -10 (1, 1, 0) - 2 (0, -11, 1) = (-10, 12, -2)$.
• Apr 3rd 2009, 10:02 PM
scorpion007
Quote:

Originally Posted by mr fantastic
So $\displaystyle c = -10 \lambda$ and $\displaystyle b = 11 \lambda$.

Could you explain just this step please? It seems to be what's giving me trouble. I don't see how you derived this. The rest makes sense.
• Apr 3rd 2009, 11:04 PM
mr fantastic
Quote:

Originally Posted by scorpion007
Could you explain just this step please? It seems to be what's giving me trouble. I don't see how you derived this. The rest makes sense.

$\displaystyle a - b = c$ .... (1)

$\displaystyle a + b = c - 11 d$ .... (2)

(1) + (2): $\displaystyle 2a = 2c - 11 d \Rightarrow 2 \lambda = 2c + 22 \lambda \Rightarrow -20 \lambda = 2c$

Substitute $\displaystyle c = -10 \lambda$ and $\displaystyle a = \lambda$ into equation (1) and solve for $\displaystyle b$.